For the imaginary unit i (where i^2 = -1), which expression is equivalent to i^233?

Introduction:
This question checks your knowledge of powers of the imaginary unit i in complex numbers. Because powers of i repeat in a simple cycle, you can evaluate very large exponents without heavy computation.

Given Data / Assumptions:

• i is the imaginary unit with i^2 = -1.
• We need to simplify i^233.
• We are working in the standard complex number system.

Concept / Approach:
Powers of i follow a cycle of length 4: i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1, and then the pattern repeats: i^5 = i, i^6 = -1, and so on. Therefore, to simplify i^233, we only need the remainder of 233 when divided by 4.

Step-by-Step Solution:
Step 1: Find 233 mod 4. Divide 233 by 4: 4 * 58 = 232 with remainder 1. So 233 = 4 * 58 + 1 and: 233 mod 4 = 1. Step 2: Use the power cycle. In general: i^(4k + 1) = i, i^(4k + 2) = -1, i^(4k + 3) = -i, i^(4k) = 1. Since 233 = 4 * 58 + 1, the exponent is of the form 4k + 1.
Therefore: i^233 = i^(4 * 58 + 1) = i.

Verification / Alternative Check:
We can verify using smaller examples: i^5 = i^(4 + 1) = i, i^9 = i^(8 + 1) = i, and so on. Every time the exponent is 1 more than a multiple of 4, the result is i. So i^233 matches this pattern.

Why Other Options Are Wrong:
1 corresponds to exponents of the form 4k, not 4k + 1.
-1 corresponds to exponents of the form 4k + 2.
-i corresponds to exponents of the form 4k + 3.
Since 233 leaves a remainder of 1 when divided by 4, only i is correct.

Common Pitfalls:
A frequent mistake is to try to compute large powers directly or to forget the exact order of the cycle. Another error is mishandling the division to find the remainder. Always reduce the exponent modulo 4 when working with powers of i.

Final Answer:
The expression i^233 is equivalent to i.